1. ## race problem

In a race of 1000 meters, A beats B by 200 meters and A beats C by 500 meters. Assuming that the contestants run at constant speeds, by how many meters does B beat C?

(A) 300 (B) 375 (C) 450 (D) 500 (E) 625

2. Originally Posted by sri340
In a race of 1000 meters, A beats B by 200 meters and A beats C by 500 meters. Assuming that the contestants run at constant speeds, by how many meters does B beat C?

(A) 300 (B) 375 (C) 450 (D) 500 (E) 625
A speed = $\frac{1000}{t}$

B speed = $\frac{800}{t}$

C speed = $\frac{500}{t}$

After t, B has 200 metres to travel, an extra quarter of his distance already travelled in t (not a fifth of 1000, it's a quarter of 800).
This will take him an extra $\frac{t}{4}$

In $\frac{t}{4}$, C will have travelled a quarter of 500,

which is 125 metres.

Therefore C still has 1000-(500-125) metres to travel, when B crosses the line.

3. Hello, sri340!

Another approach (very primitive) . . .

In a 1000-m race, $A$ beats $B$ by 200 meters and $A$ beats $C$ by 500 meters.
Assuming that the contestants run at constant speeds, by how many meters does $B$ beat $C$?

. . $(A)\;300 \qquad(B)\;375 \qquad (C)\;450 \qquad (D)\;500 \qquad (E)\; 625$

The race ended like this:
Code:
                          C           B       A
+ - - - - - - - - - o - - - - - o - - - o
0                  500         800    1000

$C$ ran 500 m in the time that $B$ ran 800 m.
C's speed is $\tfrac{500}{800} = \tfrac{5}{8}$ of $B$'s speed.

When $B$ ran his last 200 m, $C$ ran only: . $\tfrac{5}{8}(200) \,=\,125$ m.

As $B$ crosses the finish line, $C$ is at the 625-m mark.

Therefore, $B$ beats $C$ be 375 m . . . answer (B).